Pressure on circular plate (2.2 m)

The plate is not vertical because its depth range (4.5 − 2.5 = 2 m) is less than its diameter, so it is inclined. Use the formulas for an inclined plane surface.

Given: $d = 2.2$ m, $S = 1.2$ so $\rho = 1200$ kg/m³, depths of top and bottom points $= 2.5$ m and $4.5$ m.

  free surface =====================
         :  2.5 m
         :     * top point
         :    / \
         :   /   \   plate, d = 2.2 m
         :  /  G  \   inclined at theta
         :   \   /
         :    \ /
         :     * bottom point (4.5 m)

Inclination and centroid depth

$$\begin{aligned} \sin\theta &= \frac{4.5 - 2.5}{2.2} = 0.9091 \\ \bar h &= \frac{2.5 + 4.5}{2} = 3.5\ \text{m} \end{aligned}$$

Area and second moment of area

$$\begin{aligned} A &= \frac{\pi}{4}(2.2)^2 = 3.8013\ \text{m}^2 \\ I_G &= \frac{\pi d^4}{64} = \frac{\pi (2.2)^4}{64} = 1.1499\ \text{m}^4 \end{aligned}$$

Total pressure

$$\begin{aligned} F &= \rho g A \bar h = 1200 \times 9.81 \times 3.8013 \times 3.5 \\ &= 156\,622\ \text{N} = 156.62\ \text{kN} \end{aligned}$$

Centre of pressure

$$\begin{aligned} h^* &= \bar h + \frac{I_G \sin^2\theta}{A \bar h} \\ &= 3.5 + \frac{1.1499 \times 0.8264}{3.8013 \times 3.5} \\ &= 3.5 + 0.0714 = 3.571\ \text{m} \end{aligned}$$

Answer: total pressure $F \approx 156.62$ kN; centre of pressure at a depth of $3.571$ m below the free surface (on the plate's vertical diameter, below the centroid).